Compare Your Answers
Completion requirements
Identify any non-permissible values, and simplify the expressions.
-
\(\frac{{9d}}{{4f^2}} - \frac{{3d}}{{4f^2 }}\)
NPVs:
\(\begin{align}
f^2 &\ne 0 \\
f &\ne 0 \\
\end{align}\)
\[\begin{align}
\frac{{9d}}{{4f^2 }} - \frac{{3d}}{{4f^2 }} &= \frac{{9d - 3d}}{{4f^2 }} \\
&= \frac{{6d}}{{4f^2 }} \\
&= \frac{{3d}}{{2f^2 }},\thinspace f \ne 0 \\
\end{align}\]
-
\(\frac{6}{{c\left( {c + 2} \right)}} - \frac{2}{c}\)
NPVs:
\(c \ne 0\)\(\begin{align}
c + 2 &\ne 0 \\
c &\ne - 2 \\
\end{align}\)\[\begin{align}
\frac{6}{{c\left( {c + 2} \right)}} - \frac{{2\left( {\color{red}{c + 2}} \right)}}{{c\left( {\color{red}{c + 2}} \right)}} &= \frac{6}{{c\left( {c + 2} \right)}} - \frac{{2\left( {c + 2} \right)}}{{c\left( {c + 2} \right)}} \\
&= \frac{{6 - \left( {2c + 4} \right)}}{{c\left( {c + 2} \right)}} \\
&= \frac{{2 - 2c}}{{c\left( {c + 2} \right)}},\thinspace c \ne - 2,\thinspace 0 \\
\end{align}\]
-
\(\frac{{8x}}{{4x + 1}} + \frac{3}{{4x^2 + 13x + 3}}\)
NPVs:
\(4x^2 + 13x + 3 = \left( {4x + 1} \right)\left( {x + 3} \right)\)
\(\begin{align}
4x + 1 &\ne 0 \\
x &\ne - \frac{1}{4} \\
\end{align}\)\(\begin{align}
x + 3 &\ne 0 \\
x &\ne - 3 \\
\end{align}\)
\[\begin{align}
\frac{{8x}}{{4x + 1}} + \frac{3}{{4x^2 + 13x + 3}} &= \frac{{8x}}{{4x + 1}} + \frac{3}{{\left( {4x + 1} \right)\left( {x + 3} \right)}} \\
&= \frac{{\left( {8x} \right)\left( {\color{red}{x + 3}} \right)}}{{\left( {4x + 1} \right)\left( {\color{red}{x + 3}} \right)}} + \frac{3}{{\left( {4x + 1} \right)\left( {x + 3} \right)}} \\
&= \frac{{8x\left( {x + 3} \right)}}{{\left( {4x + 1} \right)\left( {x + 3} \right)}} + \frac{3}{{\left( {4x + 1} \right)\left( {x + 3} \right)}} \\
&= \frac{{8x^2 + 24x + 3}}{{\left( {4x + 1} \right)\left( {x + 3} \right)}},\thinspace x \ne - 3, \thinspace -\frac{1}{4} \\
\end{align}\]
| For further information about adding and subtracting rational expressions, see pp. 332 to 335 of Pre-Calculus 11. |